CodeForcesGym 100753F Divisions

Divisions

Time Limit: 2000ms

Memory Limit: 262144KB

This problem will be judged on CodeForcesGym. Original ID: 100753F
64-bit integer IO format: %I64d      Java class name: (Any)

 

解题：大数质因子分解

 #include <bits/stdc++.h>
 using namespace std;
 typedef long long LL;
 const int maxn = ;
 LL mul(LL a,LL b,LL mod) {
     if(!a) return ;
     return ((a&)*b%mod + (mul(a>>,b,mod)<<)%mod)%mod;
 }
 LL quickPow(LL a,LL d,LL n) {
     LL ret = ;
     while(d) {
         if(d&) ret = mul(ret,a,n);
         d >>= ;
         a = mul(a,a,n);
     }
     return ret;
 }
 bool check(LL a,LL d,LL n) {
     if(n == a) return true;
     while(~d&) d >>= ;
     LL t = quickPow(a,d,n);
     while(d < n- && t !=  && t != n-) {
         t = mul(t,t,n);
         d <<= ;
     }
     return (d&) || t == n-;
 }
 bool isP(LL n) {
     if(n == ) return true;
     if(n <  ||  == (n&)) return false;
     static int p[] = {,,,,};
     for(int i = ; i < ; ++i)
         if(!check(p[i],n-,n)) return false;
     return true;
 }
 LL gcd(LL a,LL b) {
     if(a < ) return gcd(-a,b);//特别注意，没这个TLE
     return b?gcd(b,a%b):a;
 }
 LL Pollard_rho(LL n,LL c) {
     LL i = ,k = ,x = rand()%n,y = x;
     while(true) {
         x = (mul(x,x,n) + c)%n;
         LL d = gcd(y - x,n);
         if(d !=  && d != n) return d;
         if(y == x) return n;
         if(++i == k) {
             y = x;
             k <<= ;
         }
     }
 }
 LL Fac[maxn],tot;
 void factorization(LL n) {
     if(isP(n)) {
         Fac[tot++] = n;
         return;
     }
     LL p = n;
     while(p >= n) p = Pollard_rho(p,rand()%(n-)+);
     factorization(p);
     factorization(n/p);
 }
 unordered_map<LL,LL>ump;
 int main() {
     LL x;
     srand(time());
     while(~scanf("%I64d",&x)){
         tot = ;
         if(x == ) {
             puts("");
             continue;
         }
         if(isP(x)){
             puts("");
             continue;
         }
         factorization(x);
         ump.clear();
         for(int i = ; i < tot; ++i)
             ump[Fac[i]]++;
         unsigned long long  ret = ;
         for(auto &it:ump) ret *= (it.second + );
         printf("%I64u\n",ret);
     }
     return ;
 }
 /*
 999999999999999989
 100000007700000049
 */